Bob belabored,
> BAs I said, all else being equal, roughly same work required, localized
> territory, no chance of a job being rescheduled, etc. Use a Round Robin
> approach.
For that matter, if it is an ongoing situation, where this is simply a daily
load of randomly sized tasks, round
Awesome. thank you!
On Fri, Apr 17, 2020 at 1:15 PM Alex Tweedly via use-livecode <
use-livecode@lists.runrev.com> wrote:
> Or to do the full distribution ...
>
> put Njobs div Ndrivers into tMin
> put NJobs mod Ndrivers into tExtras
>
> repeat with i = 1 to NDrivers
> if i <= tExtras
Or to do the full distribution ...
put Njobs div Ndrivers into tMin
put NJobs mod Ndrivers into tExtras
repeat with i = 1 to NDrivers
if i <= tExtras then
allocate tMin+1
else
allocate tMin
end if
end repeat
However, say your drivers are sorted alphabetically, then before
As I said, all else being equal, roughly same work required, localized
territory, no chance of a job being rescheduled, etc. Use a Round Robin
approach. Start with first worker, store as next worker, assign first job,
update next worker. On to job 2 and so forth until no more jobs. Next day
Wait, what? You didn’t mention that!
Okay we do service for copiers in the greater LA area. We assign territories
for every technician and every customer. Also, the nearest technician to any
zone may be called upon to service another technician’s copiers if the assigned
technician is
That wasn’t part of the initial problem. :-)
Bob S
> On Apr 16, 2020, at 7:30 PM, Dev via use-livecode
> wrote:
>
> Since you would probably have long trips and short trips, I would try and
> allocate on an estimated time to complete basis so that all were busy for
> about the same time.
Assign the jobs round robin if all other things are equal.
Bob S
> On Apr 16, 2020, at 7:23 PM, Skip Kimpel via use-livecode
> wrote:
>
> I have 15 "jobs" that need to be assigned to 7 "drivers". How would you
> evenly distribute the jobs to the drivers? In this case one of the drivers
>
Jerry,
You did not over simplify at all! I used your basic concept and tweaked it
and then applied it to my nested repeats and BAM... got it.
Thanks again, everybody!
SKIP
On Fri, Apr 17, 2020 at 12:28 AM Jerry Jensen wrote:
> Hi Skip,
> Forgive me if this is not the answer you seek, or an
Hi Skip,
Forgive me if this is not the answer you seek, or an oversimplification, but
there is an easy way to find if the jobs can be exactly evenly distributed,
with nothing left over:
if (tjobs mod tdrivers) = 0 then // it is evenly distrutable.
.Jerry
> On Apr 16, 2020, at 7:41 PM, Skip
Maybe I’m misunderstanding but if you just want to divide the jobs you could
have a counter for the current driver and after assigning a job you increase
the counter and wrap it when you exceed the number of drivers.
If you’re looking for a more fancy job division algorithm you should take a
I have working on creating a loop that would divide the number of jobs by
the number of users and then checking to see to see if it is evenly
divisible but that is kind of where I am stuck. Everything I have tried
thus far has proven to be unsuccessful.
SKIP
On Thu, Apr 16, 2020 at 10:36 PM
It was already mentioned somewhat, but you need some sort of weight metric for
each job. Save the “smallest” job. For each driver, assign the largest and
smallest job remaining. When done, add the weight for each driver and give the
remaining job to the one with the least work.
Thanks,
Dev, I have already sorted the jobs out by proximity so I don't need to
worry about that. Just need the distribution piece down. I agree, you
have a good point an might prove valuable in the future.
SKIP
On Thu, Apr 16, 2020 at 10:33 PM Skip Kimpel wrote:
> Haha... Let me specify, I am
Haha... Let me specify, I am looking to do this programmatically :)
SKIP
On Thu, Apr 16, 2020 at 10:31 PM Dev via use-livecode <
use-livecode@lists.runrev.com> wrote:
> Since you would probably have long trips and short trips, I would try and
> allocate on an estimated time to complete basis so
Since you would probably have long trips and short trips, I would try and
allocate on an estimated time to complete basis so that all were busy for about
the same time.
> On 16-Apr-2020, at 8:23 PM, Skip Kimpel via use-livecode
> wrote:
>
> I have 15 "jobs" that need to be assigned to 7
On Apr 16, 2020, at 7:23 PM, Skip Kimpel via use-livecode
wrote:
>
> I have 15 "jobs" that need to be assigned to 7 "drivers". How would you
> evenly distribute the jobs to the drivers?
Fire two drivers.
:_)
If they complain, fire two more, and it’s still easy . . .
I have 15 "jobs" that need to be assigned to 7 "drivers". How would you
evenly distribute the jobs to the drivers? In this case one of the drivers
would have 3 jobs while the others have two.
Obviously, the number of drivers and number of jobs would fluctuate.
This has been my mind twister for
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